By using total energy conservation
ΔKE+(ΔPE)electo+(ΔPE)gravitation=0
21mV2+(kR+yQq−kRQq)+(−mgy)=0
21mV2=mgy+kQq(R1−R+y1);V2=2gy+m2kQqR(R+y)y
V2=2y[4πϵ0R(R+y)mqQ+g]
JEE Main 2020 — Physics Electromagnetism
A solid sphere of radius R carries a charge Q+q distributed uniformly over its volume. A very small point like piece of it of mass m gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge q. If it acquires a speed ν when it has fallen through a vertical height y (see figure), then (assume the remaining portion to be spherical)

Held on 5 Sept 2020 · Verified 6 Jul 2026.
v2=y[4πϵ0R2ymqQ+g]
v2=y[4πϵ0R(R+y)mqQ+g]
v2=2y[4πϵ0(R+y)3mQqR+g]
v2=2y[4πϵ0R(R+y)mqQ+g]
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